SOLUTION: A jar contains 9 blue marbles, 6 green marbles, and 12 red marbles. How many blue marbles must be added so that the probability of choosing a blue marble at random is 3/5 ?
a 3
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-> SOLUTION: A jar contains 9 blue marbles, 6 green marbles, and 12 red marbles. How many blue marbles must be added so that the probability of choosing a blue marble at random is 3/5 ?
a 3
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Question 274826: A jar contains 9 blue marbles, 6 green marbles, and 12 red marbles. How many blue marbles must be added so that the probability of choosing a blue marble at random is 3/5 ?
a 3 b 12 c 16 d 18 Found 2 solutions by checkley77, dabanfield:Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! 9/(9+6+12)
9/27
1/3 chances of selecting a blue marble
x/(x+6+12)=3/5
x/(x+18)=3/5 cross mutiply
3x+48=5x
3x-5x=-54
2x=54
x=54
x=27 blue marbles will give you a 3/5 chances of selecting a blue marble.
Proof:
27/(27+6+12)=3/5
27/45=3/5
3/5=3/5
You can put this solution on YOUR website! Let b be the number of blue marbles that need to be added. When we start there are 9 blue marbles and 27 marbles altogether. If we add b marbles we will then have 9 + b blue marbles and 27 + b total marbles. For the probability of choosing a blue marble to be 3/5 we need to have:
(9 + b)/(27 + b) = 3/5 or
5*(9 + b) = 3*(27 + b)
Solve the above equation for b.
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